Question

Solve:

$(x+1)(x-2)(x-3)<0$

Answer 1

$(x+1)(x-2)(x-3)<0$

First take boundary points as roots i.e $x=-1,x=2,x=3$

Now consider cases

$\left(1\right)$ case $\left(a\right)$ $x<-1$

We get $x+1<0,x-2<0,x-3<0$

$\therefore (x+1)(x-2)(x-3)<0$ ------- $\left(1\right)$

We have $x+1>0,x-2<0,x-3<0$

$\therefore (x+1)(x-2)(x-3)>0$ -------$\left(2\right)$

We have $x+1>0,x+2>0,x-3<0$

$\therefore (x+1)(x-2)(x-3)<0$ --------$\left(3\right)$

Now $x>3$

$\therefore (x+1)>0(x-2)>0(x-3)>0$

$\therefore (x+1)(x-2)(x-3)>0$ -------- $\left(4\right)$

$\therefore$ from $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$

$x<-1,2<x<3$

$\therefore x\in (-\infty ,-1)\cup (2,3)$